A capacitor of 2μF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :

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20%

75%

80%

answer is D.

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Detailed Solution

Initial energy stored in capacitor 2μFUi=122(V)2=V2Final voltage after switch 2 is ONVf=C1V1C1+C2=2V10=0.2VFinal energy in both the capacitorsUf=12C1+C2Vf2 =12102V102=0.2V2So, energy dissipated = V2-0.2V2V2 x 100=80%

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